Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<iostream>
#include<vector>
#include<string>
#include<algorithm>	
#include<map>
#include<set>
#include<utility>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#include<cstdio>
#include<sstream>
#include<iomanip>
#include<assert.h>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
using namespace std;
//kaewasuretyuui
typedef long long ll;
typedef int Def;
typedef pair<Def,Def> pii;
typedef vector<Def> vi;
typedef vector<vi> vvi;
typedef vector<pii> vp;
typedef vector<vp> vvp;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef pair<Def,pii> pip;
typedef vector<pip>vip;
//#define mt make_tuple
//typedef tuple<double,int,double> tp;
//typedef vector<tp> vt;
template<typename A,typename B>bool cmin(A &a,const B &b){return a>b?(a=b,true):false;}
template<typename A,typename B>bool cmax(A &a,const B &b){return a<b?(a=b,true):false;}
const double PI=acos(-1);
const double EPS=1e-7;
Def inf=sizeof(Def)==sizeof(long long)?2e18:1e9;
int dx[]={0,1,0,-1};
int dy[]={1,0,-1,0};
class DIJ{
	public:
	struct edge{
		int to,cost;
	};
	vector<vector<edge> >G;
	int n;
	vi d;//distance
	DIJ(int size){
		n=size;
		G=vector<vector<edge> >(n);
	}
	void add_edge(int a,int b,int c){
		edge e={b,c},ee={a,c};
		G[a].pb(e);
		G[b].pb(ee);
	}
	void dij(int s){
		d=vi(n,inf);
		d[s]=0;
		priority_queue<pii>q;
		q.push(pii(0,s));
		while(!q.empty()){
			pii p=q.top();
			q.pop();
			int pos=p.second,cost=-p.first;
			if(cost>d[pos])continue;
			rep(i,G[pos].size()){
				edge e=G[pos][i];
				int to=e.to;
				int nowcost=cost+e.cost;
				if(nowcost<d[to]){
					d[to]=nowcost;
					q.push(pii(-d[to],to));
				}
			}
		}
	}
};
int main(){
	int n,m,s,t;
	cin>>n>>m>>s>>t;
	s--;t--;
	DIJ dij(n);
	rep(i,m){
		int a,b;cin>>a>>b;
		a--;b--;
		dij.add_edge(a,b,1);
		dij.add_edge(b,a,1);
	}
	dij.dij(s);
	vi d=dij.d;
	dij.dij(t);
	vi co(n);
	rep(i,n)if(dij.d[i]<n)co[dij.d[i]]++;
	ll out=0;
	rep(i,n)if(d[t]-2-d[i]<n&&d[t]-2-d[i]>=0)out+=co[d[t]-2-d[i]];
	cout<<out<<endl;
}